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olya-2409 [2.1K]
3 years ago
11

Suppose a car going 65 mph is traveling north parallel to a train going 50 mph. Once the front of the car is as far north as the

back of the train it takes another three minutes for the front of the car to be further north than the front of the train. How long is the train?​
Mathematics
1 answer:
scoray [572]3 years ago
6 0

Answer:

0.75 Miles

Step-by-step explanation:

To find the answer we have to find the difference in the distance the train and car travelled

65/60 is 1.083333 which is miles per minute then multiplied by 3 is 3.25 mile gone in 3 minutes

For the train it is 50/60 which is 0.833333 and the times 3 is 2.5 miles in three minutes

3.25 miles minus 2.5 miles is 0.75 miles so that is how long the train is

You might be interested in
I need help plz help
liq [111]

I think the answer is A

3 0
3 years ago
(for 50 points) Choose an expression that is equivalent to negative two raised to the fourth power divided by negative two raise
svp [43]

The mathematical expression that is equivalent to negative two raised to the fourth power divided by negative two raised to the second power is negative two raised to the sixth power divided by negative two raised to the fourth power; option D.

<h3>What is a mathematical expression?</h3>

A mathematical expression is an expression which uses mathematical symbols and mathematical operations to represent an idea.

Two mathematical expressions are equivalent if they have the same value

The expression that is equivalent to negative two raised to the fourth power divided by negative two raised to the second power can be written as (-2)⁴/(-2)² = 2²

An equivalent expression is negative two raised to the sixth power divided by negative two raised to the fourth power written as : (-2)⁶/(-2)⁴ = 2²

In conclusion, two mathematical expressions are equivalent if they have the same value when fully expressed.

Learn more about mathematical expression at: brainly.com/question/4344214

#SPJ1

5 0
1 year ago
To earn money for a contest, the math team pays $40 for a box of spirit banners. They sell each banner for $3.50. Write an equat
ankoles [38]

Answer:

3.5b - 40 where b is the amount of banners sold.

Step-by-step explanation:

We know the team has already spent $40, and they earn #3.5 per banner.

Hence, with b representing the amount of banners sold, the equation is 3.5b - 40

7 0
3 years ago
Solve these recurrence relations together with the initial conditions given. a) an= an−1+6an−2 for n ≥ 2, a0= 3, a1= 6 b) an= 7a
8_murik_8 [283]

Answer:

  • a) 3/5·((-2)^n + 4·3^n)
  • b) 3·2^n - 5^n
  • c) 3·2^n + 4^n
  • d) 4 - 3 n
  • e) 2 + 3·(-1)^n
  • f) (-3)^n·(3 - 2n)
  • g) ((-2 - √19)^n·(-6 + √19) + (-2 + √19)^n·(6 + √19))/√19

Step-by-step explanation:

These homogeneous recurrence relations of degree 2 have one of two solutions. Problems a, b, c, e, g have one solution; problems d and f have a slightly different solution. The solution method is similar, up to a point.

If there is a solution of the form a[n]=r^n, then it will satisfy ...

  r^n=c_1\cdot r^{n-1}+c_2\cdot r^{n-2}

Rearranging and dividing by r^{n-2}, we get the quadratic ...

  r^2-c_1r-c_2=0

The quadratic formula tells us values of r that satisfy this are ...

  r=\dfrac{c_1\pm\sqrt{c_1^2+4c_2}}{2}

We can call these values of r by the names r₁ and r₂.

Then, for some coefficients p and q, the solution to the recurrence relation is ...

  a[n]=pr_1^n+qr_2^n

We can find p and q by solving the initial condition equations:

\left[\begin{array}{cc}1&1\\r_1&r_2\end{array}\right] \left[\begin{array}{c}p\\q\end{array}\right] =\left[\begin{array}{c}a[0]\\a[1]\end{array}\right]

These have the solution ...

p=\dfrac{a[0]r_2-a[1]}{r_2-r_1}\\\\q=\dfrac{a[1]-a[0]r_1}{r_2-r_1}

_____

Using these formulas on the first recurrence relation, we get ...

a)

c_1=1,\ c_2=6,\ a[0]=3,\ a[1]=6\\\\r_1=\dfrac{1+\sqrt{1^2+4\cdot 6}}{2}=3,\ r_2=\dfrac{1-\sqrt{1^2+4\cdot 6}}{2}=-2\\\\p=\dfrac{3(-2)-6}{-5}=\dfrac{12}{5},\ q=\dfrac{6-3(3)}{-5}=\dfrac{3}{5}\\\\a[n]=\dfrac{3}{5}(-2)^n+\dfrac{12}{5}3^n

__

The rest of (b), (c), (e), (g) are solved in exactly the same way. A spreadsheet or graphing calculator can ease the process of finding the roots and coefficients for the given recurrence constants. (It's a matter of plugging in the numbers and doing the arithmetic.)

_____

For problems (d) and (f), the quadratic has one root with multiplicity 2. So, the formulas for p and q don't work and we must do something different. The generic solution in this case is ...

  a[n]=(p+qn)r^n

The initial condition equations are now ...

\left[\begin{array}{cc}1&0\\r&r\end{array}\right] \left[\begin{array}{c}p\\q\end{array}\right] =\left[\begin{array}{c}a[0]\\a[1]\end{array}\right]

and the solutions for p and q are ...

p=a[0]\\\\q=\dfrac{a[1]-a[0]r}{r}

__

Using these formulas on problem (d), we get ...

d)

c_1=2,\ c_2=-1,\ a[0]=4,\ a[1]=1\\\\r=\dfrac{2+\sqrt{2^2+4(-1)}}{2}=1\\\\p=4,\ q=\dfrac{1-4(1)}{1}=-3\\\\a[n]=4-3n

__

And for problem (f), we get ...

f)

c_1=-6,\ c_2=-9,\ a[0]=3,\ a[1]=-3\\\\r=\dfrac{-6+\sqrt{6^2+4(-9)}}{2}=-3\\\\p=3,\ q=\dfrac{-3-3(-3)}{-3}=-2\\\\a[n]=(3-2n)(-3)^n

_____

<em>Comment on problem g</em>

Yes, the bases of the exponential terms are conjugate irrational numbers. When the terms are evaluated, they do resolve to rational numbers.

6 0
3 years ago
I need help with this pls use 3.14 with or the formula with area
Leni [432]
Rectangle:
A=l(w)

A=12(6)

A=72cm^2

Semicircle:
r=6/2

r=3

A=(3.14(r)^2(1/2)

A=3.14(3)^2(1/2)

A=3.14(9)(1/2)

A=14.13cm^2

A=14.1cm^2

Total area: 72cm^2 + 14.1cm^2 = 86.1cm^2
6 0
3 years ago
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